Improper Integrals: Introduction

Recall that the definition of an integral requires the function f(x) to be bounded on the bounded
interval [a,b] (where a and b are two real numbers). It is
natural then to wonder what happens to this definition if

Case Type I: Consider the function f(x) defined
on the interval [a,b] (where a and b are real numbers). We have
two cases f(x) becomes unbounded around a or unbounded around b (see the images below)

and

For the sake of illustration, we considered a positive function. The
integral represents the area of the
region bounded by the graph of f(x), the x-axis and the lines x=a
and x=b. Assume f(x) is unbounded at a. Then the trick behind
evaluating the area is to compute the area of the region bounded by
the graph of f(x), the x-axis and the lines x=c and x=b. Then
we let c get closer and closer to a (check the figure below)

Hence we have

Note that the integral is well
defined. In other words, it is not an improper integral.
If the function is unbounded at b, then we will have

Remark. What happened if the function f(x) is unbounded at
more than one point on the interval [a,b]?? Very easy, first you
need to study f(x) on [a,b] and find out where the function is
unbounded. Let us say that f(x) is unbounded at and for
example, with . Then you must choose a
number between and (that is )
and then write

Then you must evaluate every single integral to obtain the integral
. Note that the single integrals do
not present a bad behavior other than at the end points (and not for
both of them).

Example. Consider the function defined on [0,1]. It is easy to see that f(x)
is unbounded at x = 0 and .
Therefore, in order to study the integral

we will write

and then study every single integral alone.

Case Type II: Consider the function f(x) defined
on the interval or . In other words, the
domain is unbounded not the function (see the figures below).

and

The same as for the Type I, we considered a positive function just for
the sake of illustrating what we are doing. The following picture
gives a clear idea about what we will do (using the area approach)

So we have

and

Example. Consider the function defined on . We have

On the other hand, we have

Hence we have

It may happen that the function f(x) may have Type I and Type II
behaviors at the same time. For example, the integral

is one of them. As we did before, we must always split the integral
into a sum of integrals with one improper behavior (whether Type I or
Type II) at the end points. So for example, we have

The number 1 may be replaced by any number between 0 and
since the function has a Type I
behavior at 0 only and of course a Type II behavior at .